Question

1. two bags each contain 10 balls, labelled with the positive integers from 1 to 10. pierre removes one ball from each bag. (in each bag, each ball is equally like to be chosen.) determine the probability that the product of the numbers on the two balls the chooses is divisible by 10.

Explanation:

Step1: Analyze the condition for divisibility by 10

A number is divisible by 10 if it has both 2 and 5 as factors. So, for the product of two numbers (one from each bag, each with numbers 1 - 10) to be divisible by 10, one number must be a multiple of 2 (even number) and the other must be a multiple of 5.

Step2: Count the number of multiples of 2 and 5 in each bag

- Multiples of 2 (even numbers) in 1 - 10: 2, 4, 6, 8, 10. So, there are \( n_2 = 5 \) numbers.
- Multiples of 5 in 1 - 10: 5, 10. So, there are \( n_5 = 2 \) numbers.
- Numbers that are multiples of both 2 and 5 (i.e., multiples of 10): 10. So, there is \( n_{10}=1 \) number.

Step3: Calculate the number of favorable cases

Case 1: First number is a multiple of 2 (not multiple of 10) and second is a multiple of 5 (not multiple of 10).
- Multiples of 2 not multiple of 10: \( 5 - 1 = 4 \) (2, 4, 6, 8)
- Multiples of 5 not multiple of 10: \( 2 - 1 = 1 \) (5)
- Number of cases for this sub - case: \( 4\times1 = 4 \)

Case 2: First number is a multiple of 5 (not multiple of 10) and second is a multiple of 2 (not multiple of 10).
- Multiples of 5 not multiple of 10: 1 (5)
- Multiples of 2 not multiple of 10: 4 (2, 4, 6, 8)
- Number of cases for this sub - case: \( 1\times4 = 4 \)

Case 3: First number is a multiple of 10 and second is any number (since 10 provides both 2 and 5 factors)
- Multiples of 10: 1 (10)
- Number of cases: \( 1\times10 = 10 \)

Case 4: Second number is a multiple of 10 and first is any number (since 10 provides both 2 and 5 factors)
- Multiples of 10: 1 (10)
- Number of cases: \( 10\times1 = 10 \)

But we have double - counted the cases where both are multiples of 10 (i.e., when both are 10). So, we need to subtract the over - counted cases. The number of cases where both are 10 is 1 (counted in both Case 3 and Case 4).

Now, total favorable cases:
\[ \begin{align*} &(4\times1)+(1\times4)+(1\times10)+(10\times1)- 1\\ =&4 + 4+10 + 10-1\\ =&27 \end{align*} \]

Wait, another way: The correct formula for favorable cases is:
Number of ways to choose a multiple of 2 from first bag and multiple of 5 from second bag + number of ways to choose a multiple of 5 from first bag and multiple of 2 from second bag - number of ways to choose multiple of 10 from both (to avoid double - counting)

Number of ways to choose multiple of 2 from first bag: 5, multiple of 5 from second bag: 2. So, \( 5\times2 = 10 \)
Number of ways to choose multiple of 5 from first bag: 2, multiple of 2 from second bag: 5. So, \( 2\times5 = 10 \)
Number of ways to choose multiple of 10 from both bags: 1\times1 = 1 (since there is one 10 in each bag)
So, favorable cases \( = 10 + 10-1=19 \)? Wait, no, let's use the principle of inclusion - exclusion correctly.

The correct logic: For the product \( a\times b \) to be divisible by 10, \( a \) must have a factor of 2 and \( b \) must have a factor of 5, or \( a \) must have a factor of 5 and \( b \) must have a factor of 2.

Let \( A \) be the event that the first number is even (multiple of 2) and \( B \) be the event that the second number is a multiple of 5. Let \( C \) be the event that the first number is a multiple of 5 and \( D \) be the event that the second number is even.

\( n(A)=5,n(B) = 2,n(C)=2,n(D)=5 \)

The number of elements in \( A\cap B \): \( n(A\cap B)=n(A)\times n(B)=5\times2 = 10 \)

The number of elements in \( C\cap D \): \( n(C\cap D)=n(C)\times n(D)=2\times5 = 10 \)

But the cases where both numbers are 10 (i.e., \( (10,10) \)) are counted in both \( A\cap B \) and \( C\cap D \). The number of such overlapping cases is 1 (since there is one 10 in each bag, so only one pair (10,10)).

So, by the principle of inclusion - exclusion for the union of two sets \( (A\cap B)\cup(C\cap D) \), the number of elements is \( n(A\cap B)+n(C\cap D)-n((A\cap B)\cap(C\cap D)) \)

\( n((A\cap B)\cap(C\cap D)) \) is the number of cases where first is even and multiple of 5 (i.e., 10) and second is multiple of 5 and even (i.e., 10), so it's 1 (the pair (10,10)).

So, number of favorable cases \( = 10 + 10-1=19 \)

Wait, let's list all possible pairs:

Multiples of 2: 2,4,6,8,10; Multiples of 5:5,10

Pairs where first is multiple of 2 and second is multiple of 5:
(2,5),(2,10),(4,5),(4,10),(6,5),(6,10),(8,5),(8,10),(10,5),(10,10) → 10 pairs

Pairs where first is multiple of 5 and second is multiple of 2:
(5,2),(5,4),(5,6),(5,8),(5,10),(10,2),(10,4),(10,6),(10,8),(10,10) → 10 pairs

But (10,10) is counted twice, so total favorable pairs \( = 10 + 10 - 1=19 \)

Step4: Calculate the total number of possible cases

Since we are choosing one ball from each of two bags with 10 balls each, the total number of possible cases is \( N = 10\times10=100 \)

Step5: Calculate the probability

The probability \( P \) is the number of favorable cases divided by the total number of cases. So, \( P=\frac{19}{100}=0.19 \)

Wait, let's re - examine the favorable cases.

A product \( a\times b \) is divisible by 10 iff \( a \) has a factor of 2 and \( b \) has a factor of 5, or \( a \) has a factor of 5 and \( b \) has a factor of 2.

Let's count the number of pairs where:

- \( a \) is even (2,4,6,8,10) and \( b \) is a multiple of 5 (5,10):
- When \( a\in\{2,4,6,8\} \) (4 numbers) and \( b = 5 \) (1 number): \( 4\times1 = 4 \)
- When \( a\in\{2,4,6,8\} \) (4 numbers) and \( b = 10 \) (1 number): \( 4\times1 = 4 \)
- When \( a = 10 \) (1 number) and \( b\in\{5,10\} \) (2 numbers): \( 1\times2 = 2 \)
- Sub - total for \( a \) even and \( b \) multiple of 5: \( 4 + 4+2=10 \)

- \( a \) is a multiple of 5 (5,10) and \( b \) is even (2,4,6,8,10):
- When \( a = 5 \) (1 number) and \( b\in\{2,4,6,8\} \) (4 numbers): \( 1\times4 = 4 \)
- When \( a = 5 \) (1 number) and \( b = 10 \) (1 number): \( 1\times1 = 1 \)
- When \( a = 10 \) (1 number) and \( b\in\{2,4,6,8,10\} \) (5 numbers): \( 1\times5 = 5 \)
- Sub - total for \( a \) multiple of 5 and \( b \) even: \( 4 + 1+5 = 10 \)

Now, the overlapping case (when \( a = 10 \) and \( b = 10 \)) is counted in both sub - totals. So, total favorable cases \( = 10+10 - 1=19 \) (since (10,10) is counted once in the first sub - total and once in the second, we subtract 1 to count it only once).

Total number of possible pairs: \( 10\times10 = 100 \)

So, the probability \( P=\frac{19}{100}=0.19 \)

Answer:

The probability that the product of the numbers on the two balls is divisible by 10 is \(\frac{19}{100}\) (or 0.19).